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# HW6sol - ORIE 3500/5500 Fall Term 2008 Assignment...

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ORIE 3500/5500 Fall Term 2008 Assignment 6-Solution 1. The joint density of X, Y, Z is f X,Y,Z ( x, y, z ) = f X ( x ) f Y ( y ) f Z ( z ) = 1 if 0 x, y, z 1 0 otherwise . (a) We have P ( X > Y + Z ) = [ x>y + z ] f X,Y,Z ( x, y, z ) dxdydz = 1 0 1 - z 0 1 y + z 1 dxdydz = 1 0 1 - z 0 (1 - y - z ) dydz = 1 0 (1 - z - (1 - z ) 2 / 2 - z (1 - z )) dz = 1 0 (1 - z ) 2 / 2 dz = (1 / 2) 1 0 w 2 dw = 1 / 6 . (b) We have P ( Y < ZX ) = [ y<zx ] f X,Y,Z ( x, y, z ) dydzdx = 1 0 1 0 zx 0 1 dydzdx = 1 0 1 0 zxdzdx = 1 0 x (1 / 2) dx = 1 / 4 . 1

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(c) Let W = X + Y . for w [0 , 2], F W ( w ) = P ( W w ) = min( w, 1) 0 min( w - x, 1) 0 1 .dydx = w 0 ( w - x ) 0 dx if w < 1 w - 1 0 1 dx + 1 w - 1 ( w - x ) dx if w 1 = w 2 / 2 if w < 1 2 w - 1 - w 2 / 2 if w 1 and hence f W ( w ) = F W ( w ) = w if 0 w 1 2 - w if 1 w 2 0 otherwise . 2. We know f Y ( y ) = 1 if 0 y 1 0 otherwise , f X | Y ( x | y ) = 1 2(1 - y ) if - (1 - y ) x (1 - y ) 0 otherwise . (a) We have f X ( x ) = f Y ( y ) f X | Y ( x | y ) dy = 1 - x 0 1 2(1 - y ) dy if x > 0 1+ x 0 1 2(1 - y ) dy if x 0 = - (1 / 2) log | x | , if - 1 x 1 0 otherwise . (b) Since f X ( x ) is symmetric about 0, E ( X ) = 0. var ( X ) = E ( X 2 ) = - (1 / 2) 1 - 1 x 2 log | x | dx = - 1 0 x 2 log( x ) dx = - log( x ) · ( x 3 / 3) | 1 0 - 1 0 (1 /x )( x 3 / 3) dx = - [0 - 1 / 9] = 1 / 9 .
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